CHEMISTRY TEXTBOOK

Can you recall?

What are the products of reaction

of

i. CH 3 - CO - CH 3 with NH 2 - OH?

ii. CH 3 - CHO with HCN?

iii. CH 3 - OH with CH 3 - CO - O - CO- CH 3?

b. Prepartion of glucose from starch :

Commercially glucose is obtained by
hydrolysis of starch by boiling it with dilute
sulfuric acid at 393K under 2 to 3 atm
pressure.

(C 6 H 10 O 5 )n + n H 2 O H

⊕

393K, 2-3 atm n C^6 H^12 O^6

(Starch) (Glucose)

14.2.4 Structure and properties of glucose

Glucose has an aldohexose structure.
In other words, glucose molecule contains
one aldehydic, that is, formyl group and the
remaining five carbons carry one hydroxyl
group (-OH) each. The six carbons in glucose
form one straight chain. This aldohexose
structure of glucose was established on the
basis of the following chemical properties.

1. Molecular formula of glucose was found
   to be C 6 H 12 O 6 , on the basis of its elemental
   compostion and colligative properties.


2. The six carbons in glucose molecule form
   a straight chain. This was inferred from
   the following observation : Glucose gives
   n-hexane on prolonged heating with HI.

CHO - (CHOH) 4 - CH 2 OH
(Glucose)
HI, ∆ CH 3 - (CH 2 ) 4 - CH 3

(n-Hexane)

3. Glucose molecule contains one carbonyl
   group. This was inferred from the observation
   that glucose forms oxime by reaction with
   hydroxylamine and gives cyanohydrin on
   reaction with hydrogen cyanide.
   4. The carbonyl group in glucose is in the
   form of aldehyde. This was inferred from the
   observation that glucose gets oxidised to a six
   carbon monocarboxylic acid called gluconic
   acid on reaction with bromine water which is
   a mild oxidizing agent.

Br 2 water

(O)

CHO

(CHOH) 4

CH 2 OH

(Glucose)

COOH

(CHOH) 4

CH 2 OH

(Gluconic acid)

Problem 14.1 :

An alcoholic compound was found to have

molcular mass of 90 u. It was acetylated.

Molecular mass of the acetyl derivative

was found to be 174 u. How many alcoholic

(-OH) groups must be present in the original

compound?

Solution : In acetylation reaction H atom

of an (-OH) group is replaced by an acetyl

group (-COCH 3 ). This results in an increase

in molcular mass by [(12+16+12+3×1)-1],

that as, 42 u.

In the given alcohol,

increase in molecular mass = 174 u - 90 u

= 84 u

∴ Number of -OH groups = 84 u42 u = 2

5. Glucose contains five hydoxyl groups :
   This was inferred from the observation that
   Glucose reacts with five moles of acetic
   anhydride to form glucose pentaacetate. As
   glucose is a stable compound, it was further
   inferred that the five hydroxyl groups are
   bonded to five different carbon atoms in
   glucose molecule.

NH^2

OH HCN

CH

(CHOH) 4

CH 2 OH

(Cyanohydrin)

OH

CH=N-OH CN

(CHOH) 4

CH 2 OH

(Oxime)

CHO

(CHOH) 4

CH 2 OH

-H^2 (Glucose)

O